On the Ext

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arXiv:1908.06529v1 [math.FA] 18 Aug 2019

On the Ext -problem for Hilbert spaces

Félix Cabello Sánchez, Jesús M.F. Castillo, Willian H. G. Corrêa, Valentin Ferenczi, and Ricardo Garc´ıa

Abstract. We show that Ext²(ℓ₂, ℓ₂)6= 0 in the category of Banach spaces. This solves a sharpened version of Palamodov’s problem and provides a solution to the second order version of Palais problem. We also show that Ext²(ℓ₁,K)6= 0 in the category of quasi Banach spaces which solves the four-space problem for local convexity.

1. Introduction

This paper is devoted to solving the main problem left open in [5] and [8]: Is Ext²(ℓ2, ℓ2) = 0? To answer the question means to decide whether every four-term exact sequence of Banach spaces and operators

0 −−−→ ℓ2 −−−→ E −−−→ F −−−→ ℓ2 −−−→ 0

is trivial, that is, equivalent to zero in Ext²(ℓ2, ℓ2). The difficulties to do this are of two kinds:

(a) How to construct such an object.

(b) How to decide if it is trivial or not.

What makes (a) so difficult is the perfect homogeneity of Hilbert space. General methods provide examples of Banach spacesX, Y so that Ext²(X, Y)6= 0 usually appealing to some incomparability property between projective and injective presentations, or between X and Y, or their (complemented) subspaces or quotients. In the case of ℓ₂ nothing works:

the space is self-similar in any way, as well as its subspaces, quotients or dual. Moreover, attempts to use the standard reduction of degree technique fail because very few things are known either aboutK¹(ℓ2), the kernel of a quotient map ℓ1 −→ ℓ2 or its dual space ℓ∞/ℓ2. In particular, no known technique allows one to obtain a nontrivial elements in either Ext(K¹(ℓ2), ℓ2) or Ext(ℓ2, ℓ∞/ℓ2). In turn, what makes (b) difficult is that, when n≥2, the equivalence relation in Extⁿ is not easy to handle, to say the least.

2010 Mathematics Subject Classification. 46B25,46M18.

The research of the first, second and fifth authors was supported by Project MINCIN MTM2016- 76958-C2-1-P and Project IB16056 de la Junta de Extremadura. The research of the third author is part of the FAPESP project 2016/25574-8 and has been supported by CAPES, PDSE program, grant 88881.134107/2016-0. The fourth author was supported by CNPq, grant 303034/2015-7 and Fapesp, grant 2016/25574-8.

1

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In spite of all this we will show that Ext²(ℓ2, ℓ2) 6= 0 by means of a counterexample which is optimal in every sense: First, because it does have the simplest possible form

(1) 0 ^//ℓ2 //Ze2 //

&&

▲▲

▲▲ Z2 //ℓ2 // 0

ℓ2

88r

rr rr r

of two twisted Hilbert sequences spliced. Second, because the short exact sequence on the right 0 −→ ℓ2 −→ Z2 −→ ℓ2 −→ 0 is Kalton-Peck’s celebrated example [20], while the one on the left 0 −→ ℓ2 −→ Ze2 −→ ℓ2 −→ 0 is a certain vector-valued form of the Kalton-Peck sequence. And third, because the construction preserves unconditionality in the sense that Diagram 1 naturally belongs to the category of Banach modules over ℓ∞, which considerably expands its range of applications.

After all this propaganda, let us explain the organization of the paper and highlight the main results. Section 2 is preliminary. It contains the necessary background about Ext = Ext¹, quasilinear maps and centralizers, as well as some ancillary material on Ext². Nonlinear maps find their way here to allow the construction and study of the short exact sequences later spliced to get the required four-term sequence. The basic idea is that each exact sequence 0 −→ Y −→ Z −→ X −→ 0 corresponds to a quasilinear map Φ :X −→Y, out from which the middle spaceZ can be reconstructed. The price we pay for this simplification is that our default setting has to be the category of quasi Banach spaces, even if we are mostly interested in Banach spaces.

Section 3 contains the main criteria for the (non) triviality of sequences in Ext². The equivalence relation of Extⁿ is quite elusive when n ≥ 2. However, a small symmetry miracle occurs at n = 2 which provides a criterion, both visual from the diagrammatic point of view and operative from the quasilinear point of view. We then develop some ad-hoc estimates which not only render the basic criterion manageable, but even suggest the right form of the counterexamples.

Armed with these tools we tackle in Section 4 the task of proving the nontriviality of (1). This is achieved by means of a selective elimination of the symmetries of Z2 and a combinatorial argument involving partitions.

Section 5 contains a number of applications, ranging from operator theory to the (non) vanishing of Ext² both in Banach modules and quasi Banach spaces, including:

• That Ext²(X, Y) 6= 0 for all Banach spaces X and Y containing ℓⁿ₂ uniformly complemented.

• A proof that every four-term exact sequence obtained by splicing two copies of the same centralizer onℓp is trivial in Ext²(ℓp, ℓp); and that the same occurs with the Enflo-Lindenstrauss-Pisier quasilinear map [10], the first quasilinear map appearing in Banach space theory, who is not a centralizer.

• Some remarks on the Yoneda product Ext(ℓp, ℓp)×Ext(ℓp, ℓp) −→ Ext²(ℓp, ℓp) for module extensions.

• A proof that Ext²(ℓp, ℓp) is not zero in the category of quasi Banach modules over ℓp.

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• A proof that Ext²(ℓ1,K) is not zero in in the category of quasi Banach spaces.

Precisely, a nontrivial exact sequence of quasi Banach spaces 0−→K−→R −→

B −→ℓ₁ −→0 where K is the ground field andR is Ribe’ space [25].

The problem addressed in this paper can be regarded as a sharpened version of Palam- odov [24, Problem 6]: is Ext²(·, E) = 0 for any Fr´echet space? Counterexamples were provided by Wengenroth [27, pp. 177–178], and then in [8]. In the first case the example has the form

0 ^//U^∗ ^// L∞ //

ı◆^∗◆◆◆◆''

◆ L1 //U ^// 0

ℓ2 ı

88q

qq qq q

whereı is an isomorphic embedding, while in the second case the example is

0 ^//ℓ2 //X ^//

&&

▼▼

▼▼ L1 // U ^// 0

ℓ2 ı

88q

qq qq q

where 0−→ℓ2 −→X −→ℓ2 −→0 is a nontrivial twisted Hilbert space. In both cases an uncontrolled space U =L1/ı[ℓ2] appears that cannot be reduced to anything reasonable, let alone to a Hilbert space. Thus, the question of the vanishing of Ext²(ℓ2, ℓ2) remained untractable by general methods.

The introduction of homological methods in Banach space theory was fueled by the attempt to solve what is known as Palais problem: does there exist a twisted Hilbert space that is not (isomorphic to) a Hilbert space? The existence of such object was first proved by Enflo, Lindenstrauss and Pisier [10], but the construction which is of paramount importance to our purposes is that of the Kalton-Peck space Z₂ appeared soon later in [20]. Thus, our proof that Ext²(ℓ2, ℓ2) 6= 0 can be viewed as a solution to the “second order” Palais problem. The closely related fact that Ext²(ℓ₁,K) 6= 0, within the category of quasi Banach spaces, can be viewed as an optimal solution of the

“four-spaces problem” for local convexity. The classical “three-space problem” for local convexity was solved independently and almost simultaneously by Roberts [26], Ribe [25]

and Kalton [14] in the late seventies.

2. Preliminaries

Although most of the time we deal with Banach spaces which are in fact quite similar to Hilbert spaces, the natural setting for our study is the category of quasi Banach spaces and linear bounded operators, that we denote by Q when necessary. The subcategory of Banach spaces is denoted byB. Our main examples share a certain unconditional structure that makes them into quasi Banach modules over the Banach algebraℓ_∞in a natural way and so it will also be convenient to consider the category of quasi Banach ℓ∞-modules and linear bounded homomorphisms as well. We are aware that this approach may annoy some readers, but we hope they will feel more comfortable as the paper progresses. By the way of compensation we have made a great effort to present all our results avoiding interpolation theory (compare, when it corresponds, with [5]).

A presentation of the basic elements of (quasi) Banach space theory, including quasilinear maps and twisted sums, very akin to the approach of this paper, can be found

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in [17]. The article [19] explains the connections between centralizers and interpolation theory and many other things.

2.1. Quasi Banach spaces and modules. A quasinorm on a linear space X is a functionk · k:X −→[0,∞) satisfying the following conditions:

• kλxk=|λ|kxk for every x∈X and every scalarλ; kxk= 0 if and only if x= 0.

• There is a constant C such thatkx+yk ≤C kxk+kyk

for every x, y ∈X.

A quasinormed space is a linear spaceXequipped with a quasinorm. Such a space carries a linear topology for which the unit ball {x : kxk ≤ 1} is a neighbourhood of zero. If the resulting topological vector space is complete we call it a quasi Banach space. A quasinormed module over a Banach algebra A is a quasinormed space X together with a jointly continuous outer product A × X −→ X satisfying the traditional algebraic requirements. A quasi Banach module is a complete quasinormed module.

If X and Y are quasinormed modules over A, an homomorphism f : X −→ Y is an operator such thatf(ax) =a(f(x)) for every x∈X and a∈A.

The only Banach algebra that we need in this paper is the algebra ℓ∞ of bounded functionsa:N−→K endowed with the sup norm.

2.2. Homology. We assume from the reader some acquaintance with the basic elements of homology as expounded in the classic books [13, 22] or in the functional analysis-oriented notes [11].

Let us however fix the notation by recalling a few definitions. A (finite or infinite) sequence of quasi Banach spaces and operators

· · · ^//E_n−1 ^uⁿ⁻¹ ^//E_n ^uⁿ ^// E_n+1 ^//· · ·

is exact when the kernel of each operator agrees with the image of the preceding one:

un−1[En−1] = kerun for every n under consideration.

Let us first consider short exact sequences, that is, exact sequences of the form

(2) 0 ^//Y ^ı ^// E ^π ^//X ^//0

The reason behind this notation is that one treats the end spaces X and Y as “fixed”

while the middle space E is considered “variable”. In this setting the sequence 0 ^//Y ^//F ^// X ^//0 is equivalent to (2) if there is an operator u making the following diagram commutative:

0 ^//Y ^// E ^//

u

X ^//0

0 ^//Y ^//F ^//X ^//0

This is really an equivalence relation since a fortunate combination of algebra (the five- lemma, see [13, Lemma 1.1] or [11, Corollary 3.23]) and topology (the open mapping theorem) guarantees thatuis a linear homeomorphism. Then Ext(X, Y) is defined as the set of short exact sequences of the form (2) modulo equivalence. This set can be given a natural (= functorial) linear structure via pull-back and push-out constructions, called Baer’

sum, that the reader can see in [13, Chapter IV,§9] or [11, Chapter 6]. The zero element of Ext(X, Y) is the (class of the) trivial sequence 0 ^//Y ^//Y ⊕X ^// X ^//0

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with the obvious operators. It turns out that (2) is trivial if and only if it splits, in the sense that there is an operator P :E −→Y such that P ı=I_Y or, equivalently, there is an operator J :X −→E such thatπ J =I_X.

The definition of Ext²(X, Y) starts the same. Given two exact sequences 0−→Y −→

E1 −→E2 −→X −→0 and 0−→Y −→F1 −→F2 −→X −→0, denoted by E and F, respectively, we write E →F or F ←E if there is a commutative diagram

0 ^//Y ^// E1 //

u₁

E2 //

u₂

X ^// 0

0 ^//Y ^//F1 //F2 //X ^// 0

The maps u1 and u2 are not longer isomorphisms and we obtain just a partial order for four-term sequences. Nevertheless, this partial order generates an equivalence relation by declaringE and F equivalent (written E ∼F) if there is a finite chain (C_i)1≤i≤n so that (3) E →C₁ ←C₂ → · · · ←C_n→F.

Although it will not be used in this paper, it can be remarked that, as can be seen in [11, Corollary 6.40], two links are enough. The space Ext²(X, Y) is the set of equivalence classes of four-term exact sequences 0 −→ Y −→ E1 −→ E2 −→ X −→ 0. It carries a linear structure whose zero element is the class of the trivial sequence

0 −−−→ Y −−−→^I Y −−−→⁰ X −−−→^I X −−−→ 0

Regarding the subcategories of Banach spaces and quasi-Banach ℓ∞-modules, one can formalize the corresponding definitions of

Ext_B(X, Y), Ext²_B(X, Y), Ext_ℓ∞(X, Y), Ext²_ℓ∞(X, Y)

in the obvious way. Observe that if a given exact sequence of Banach spaces is zero in Ext²_B(X, Y), then so is in Ext²(X, Y), and the same happens for quasi Banach modules.

The converse seems to be unknown for Banach spaces and known to be false for modules, even for short exact sequences. The reason is that some of the links (spaces or arrows) in (3) might not be in the corresponding subcategory.

2.3. Quasilinear maps. The study of short exact sequences of (quasi) Banach spaces is greatly simplified by the use of quasilinear maps. A map Φ :X −→Y acting between quasinormed spaces is quasilinear if it is homogeneous (Φ(λx) = λΦ(x) for every x∈ X and λ∈K) and satisfies an estimate

kΦ(x+y)−Φ(x)−Φ(y)k ≤Q kxk+kyk

for some constantQ≥0 and all x, y ∈X. When necessary, we denote byQ(Φ) the least possible constant for which the preceding inequality holds.

These maps make their way into the homology of (quasi) Banach spaces because of the following construction. AssumeX andY are quasi Banach spaces and that Φ :X0 −→Y is quasilinear, whereX0 is a dense subspace ofX. Then the functional

k(y, x)k^Φ =ky−Φ(x)k+kxk

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is a quasinorm on the product space Y ×X₀. If we denote by Y ⊕Φ X₀ the resulting quasinormed space, then we have an exact sequence

0 ^//Y ^ı ^// Y ⊕^ΦX0 π

// X0 // 0,

whereı(y) = (y,0) andπ(y, x) =x. It is clear that bothıpreserves the quasinorms, while π maps the unit ball of Y ⊕^ΦX0 onto that of X0. Thus, ifZ(Φ) denotes the completion of Y ⊕^ΦX0, then π extends to a quotient map Z(Φ) −→ X that we call again π which yields the exact sequence of quasi Banach spaces

0 ^//Y ^ı ^// Z(Φ) ^π ^// X ^//0,

called, for good reason, the sequence induced by Φ. All short exact sequences of quasi Banach spaces arise in this way, up to equivalence, although we will not use this fact.

We say that two quasilinear maps Φ,Ψ :X₀ −→Y are strongly equivalent if they induce equivalent quasinorms onY ×X₀, equivalently, if there is a constantK such thatkΨ(x)− Φ(x)k ≤Kkxk for all x∈X0.

This applies, in particular, when X and Y are Banach spaces. One however must be aware that there are short exact sequences 0 ^// Y ^//Z ^// X ^//0 in which X and Y are Banach spaces but Z is just a quasi Banach space. This cannot occur when the quotient space is B-convex, something that all the spaces ℓp are for 1< p < ∞, and we have Ext(X, Y) = Ext_B(X, Y). These technical points will play a secondary role later.

2.4. Centralizers. These are a very special type of quasilinear maps that have to do with module structures. Centralizers have their own philosophy which does not fit exactly into the general framework of quasilinear maps as expounded in the preceding Section;

see [19, Section 8]. As the only algebra that plays a role in our exposition is ℓ∞, we can provide the reader with the minimal background one needs to understand the paper right now.

By a (quasinormed) sequence space we understand a linear space X of functions x:N−→K equipped with a quasinorm k · k such that:

• The finitely supported functions are dense in X.

• If |y| ≤ |x| and x∈X, theny ∈Y and kyk ≤ kxk.

Ifei is the function that takes the value 1 atiand vanishes elsewhere we may also require thatke_ik= 1 for everyi∈N. IfXis complete we call it a quasi Banach sequence space and we call it a Banach sequence space when the quasinorm is a norm. The easiest examples are the spaces ℓp for 0 < p < ∞. These are Banach spaces when p ≥ 1. According to this, ℓ∞ itself is not a sequence space, but c0 is. A sequence space is an ℓ∞-module under the pointwise product: if x ∈X and a ∈ ℓ∞, then ax ∈X and kaxk ≤ kak^∞kxk. From now on module structures refer to ℓ∞ unless otherwise stated and we denote by X⁰ the subspace of finitely supported sequences ofX.

Definition2.1. A centralizer is a homogeneous mapping Φ :X −→Y acting between quasinormed modules that obeys an estimate of the form

kΦ(ax)−aΦ(x)k ≤Ckak^∞kxk, for some constantC and all a∈ℓ∞, x∈X.

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Our interest in centralizers stems from the fact that if X and Y are quasi Banach modules, X0 is a dense submodule of X and Φ : X0 −→ Y is a quasilinear centralizer, then the product a(y, x) = (ay, ax) makes Y ⊕^Φ X0 into a quasinormed module (and so Z(Φ) is a quasi Banach module) and the induced sequence lives in the category of quasi Banach modules. As a partial converse, ifX is a quasi Banach sequence space, then each short exact sequence of quasi Banach modules 0−→Y −→Z −→X −→0 arises, up to equivalence, from a quasilinear centralizer Φ :X⁰ −→Y. We have the following remark about “automatic quasilinearity” of centralizers.

Lemma 2.2. Every centralizer defined on a quasinormed sequence space is quasilinear.

2.5. Kalton-Peck maps and their chunked versions. The most famous quasilinear maps were introduced by Kalton-Peck in [20] and they are actually centralizers.

Letϕ :R⁺ −→K be a Lipschitz function such that ϕ(0) = 0. Then, for 0 < p <∞, the map Ω^ϕ_p :ℓ⁰_p −→ℓp defined by

(4) Ω^ϕ_p =x ϕ

p

2log kxk

|x|

is a (quasilinear) centralizer whose constantsC Ω^ϕ_p

andQ Ω^ϕ_p

depend only onpand the Lipschitz constant ofϕ. We now introduce the second (type of) centralizer that we need to carry out the proof of the main result. As all centralizers do, it stems from complex interpolation theory (see [9, Theorem 3.2]), but we will consider a direct approach which does not require any previous knowledge on interpolation.

Let (A_i) be a partition ofNthat we consider fixed in all what follows. Fix 0< p <∞. Each x ∈ ℓp can be written as x = P

ixi, where the i-th summand is xi = x1Ai, with kxk^p = P

ikxik^pp

1/p

. Using this notation, one has:

Lemma 2.3. The map Ωe^ϕ_p :ℓ⁰_p −→ℓ_p defined by

(5) Ωe^ϕ_p(x) =X

i

x_iϕ p

2log kxk kxik

is a centralizer.

Proof. Pick x∈ℓ⁰_p and a∈ℓ∞ and set ai =a1Ai. Then Ωe^ϕ_p(ax) =X

i

aixiϕ p

2log kaxk kaixik

, aΩe^ϕ_p(x) =X

i

aixiϕ p

2log kxk kxik

.

Now,

kΩe^ϕ_p(ax)−aΩe^ϕ_p(x)k^p = X

i

kaixik^p ϕp

2log kaxk ka_ix_ik

−ϕp

2log kxk kx_ik

p!1/p

≤ X

i

kaik^p∞kxik^pp

ϕp

2log kaxk ka_ix_ik

−ϕp

2log kxk kx_ik

p!1/p

≤C(Ω^ϕ_p)kak^∞kxk^p,

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where the last inequality is just the centralizer estimate of the Kalton-Peck map Ω^ϕ_p applied to the sequences (kaik^∞)i and (kxik^p)i. The choice of the function ϕ(t) = t in (4) and (5) is especially rewarding and we simply write Ωp and Ωep for these maps. When p= 2 we will even omit the subscript. We also write Zp(ϕ) = Z(Ω^ϕ_p) and Zep(ϕ) = Z(Ωe^ϕ_p). If ϕ is the identity on R₊ we just write Zp and Zep.

2.6. Mirrored centralizers. One of the core ideas in the widely ignored memory [15] is that centralizers never walk alone. To give shape to this affirmation, let us recall a result from the even more ignored [1, Lemma 5(a) and Corollary 3]:

Lemma 2.4. Let 0< q < p <∞. IfΦp is a centralizer on ℓp, then the map defined by Φ_q(f) =u|f|^q/rΦ |f|^q/p

for q⁻¹ =r⁻¹ +p⁻¹ is a centralizer on ℓq. All centralizers on ℓq arise in this way, up to strong equivalence.

We will thus say that Φq is the reflection of Φp in ℓq. As the reader may guess, there is a reason behind the slightly eccentric presentation of the centralizers of the preceding Section: parameters have been assigned in such a way that if q < p, then Ω^ϕ_q is the reflection of Ω^ϕ_p in ℓ_q, and the same occurs to their chunked versions. This will be used in Sections 5.3 and 5.6.

3. Splitting criteria

Suppose we are given two quasi Banach spaces X and Y and that we want to construct a four-term exact sequence 0 ^// Y ^//E1 //E2 //X ^//0 . The simplest (and unique) way to achieve this is to pick another space E, then construct two short exact sequences

0 ^//Y ^ı ^//E1 π

//E ^//0 (E₁)

0 ^//E ^ ^//E2

̟ //X ^//0 (E₂)

and splice them through E to get 0 ^//Y ^// E1

 π //

&&

▼▼

▼▼ E2 //X ^//0 (E)

A

88q

qq qq q

Our first task is to decide when the resulting sequence is trivial in Ext²(X, Y). We have the following criterion which is both visual from the diagrammatic side and transparent from the quasilinear point of view.

Lemma 3.1. Let (E₁) and (E₂) be two short exact sequences of quasi Banach spaces.

The following statements are equivalent:

(a) The spliced sequence (E) is trivial in Ext²(X, Y).

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(b) The sequence (E) splits, that is, there exists a quasi Banach space and a commu- tative diagram

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0 0

 y

Y Y

 y

0 −−−→ E1 −−−→ −−−→ X −−−→ 0

 y

0 −−−→ E −−−→ E2 −−−→ X −−−→ 0

 y

0 0

with exact rows and columns.

Moreover, if the sequence(E₁) is induced by the quasilinear map Φ :E −→Y, then these are equivalent to:

(c) The map Φ :E −→Y has a quasilinear extension to E2.

If the sequence(E₂) is induced by the quasilinear map Ψ :X −→E, then (a) and (b) are equivalent to:

(d) The map Ψ :X −→E has a quasilinear lifting to E₁.

Proof. We begin with the implication (b) =⇒ (a). Observe that if there is a diagram as in (b) then there is a commutative diagram

0 ^//Y ^//Y ⊕E₁ ^//

((❘❘❘❘❘❘

// X ^//0

E1

77♣

♣♣

♣

0 ^//Y ^//E1 //

))

❘❘

❘ E2 // X ^//0

E

77♦

♦♦

♦

It follows that the lower sequence is trivial since it is equivalent to the upper one and the following diagram is commutative:

0 ^//Y ^// Y ⊕E1 //

//X ^// 0

0 ^//Y Y ⁰ ^//X X ^// 0

We now stablish the implication (a) =⇒ (b). As the zero sequence has the property required in (b) with = Y ⊕ X, it is all a matter of showing that that property is preserved in the successive steps that define the equivalence of Ext²(X, Y).

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So, assume we have a commutative diagram

0 ^//Y ^//E1 //

''◆◆◆◆◆◆ E2 //

X ^//0 E

77♣

♣♣

♣

0 ^//Y ^//F1 //

''

◆◆

◆◆ F2 // X ^//0

F

77♣

♣♣

♣

and that the sequence ofE’s fits in a diagram as (6). Then one can form the commutative diagram

(7) Y

PP PP PP P

Y

PP PP PP P

Y

E1

''◆◆◆◆◆◆ //

//

''

❖❖

❖❖ X

▼▼

F1 //

PO ^//

X

E

''

◆◆

◆ //E2 //

''

❖❖

❖❖ X

▼▼

F ^//F2 //X

where the push-out refers to the inner horizontal square E1 −−−→

 y

 y F1 −−−→ PO

and the arrows beginning at PO are defined by the universal property of the push-out contruction.

By categorical duality, that is, by “reversing the arrows” and taking pull-back instead of push-out when necessary, one sees that as long as one has a commutative diagram of exact sequences

0 ^//Y ^//E1 //

''

◆◆

◆◆ E2 // X ^//0

E

77♣

♣♣

♣

0 ^//Y ^//F1 //

OO ''◆◆◆◆◆◆ F2 //

OO

X ^//0 F

77♣

♣♣

♣

OO

and theE’s satisfy condition (b), then so the F’s do.

(b) =⇒ (c) This is immediate ... if one is acquainted with the connection between quasilinear maps and short exact sequences. Assume E1 = Y ⊕^ΦE, where Φ : E −→ Y

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is quasilinear and let us draw the hypothesized commutative diagram

(8) Y

ı

Y

I

Y ⊕ΦE ^J ^//

π

//

Q

X

E ^ ^// E2 //X

in which the 0’s have been omitted. Consider the mape ∈E 7−→(0, e)∈Y ⊕ΦE. This is a linear (probably unbounded) section of the quotient mapπ. Since all exact sequences of linear spaces are trivial in the purely algebraical sense this map can be “extended” to a linear section of Q. Precisely, there is a linear map L:E2 −→ such that

• L((e)) =J(0, e) for every e∈E.

• P ◦L is the identity on E2.

In a similar vein, consider the mapping e ∈ E 7−→ (Φ(e), e) ∈ Y ⊕^Φ E, which is a (homogeneous) bounded section ofπand “extend” it to a (homogeneous) bounded section of Q, that is, a homogeneous bounded B :E₂ −→such that

• B((e)) =J(Φ(e), e) for every e∈E.

• B◦L is the identity on E2.

It is clear that the differenceB−L mapsE2 to kerQ=I[Y] and so Γ =I⁻¹◦(B−L) is a quasilinear extension of Φ.

The implication (c) =⇒(b) is obvious: if Γ :E2 −→Y is a quasilinear map extending Φ, set=Y ⊕ΓE₂, put the obvious maps and check.

(b) =⇒ (d) Assume E2 =E ⊕^ΨX, where Ψ : X −→ E is quasilinear. The relevant diagram is now

(9) Y

ı

Y

I

E1 J

//

π

//

Q

X

E ^^//E⊕^ΨX ^//X

LetB :X −→ be a homogeneous bounded map such thatQ(B(x)) = (Ψ(x), x) for all x∈X and L:X −→ be a linear map such thatQ(L(x)) = (0, x) for all x∈X. Then Λ =J⁻¹◦(B−L) is the required lifting of Ψ, that is,π◦Λ = Ψ.

One disavantage of working with quasilinear maps is that, as a rule, they cannot be explicitly defined on the whole space which is aimed to be the quotient of the resulting short exact sequence, but only on a dense subspace, and so we need to adapt our criteria to this setting.

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Let X, Y, E be quasi Banach spaces and let Ψ : X₀ −→ E and Ψ : E₀ −→ Y be quasilinear maps, where X0 and E0 are dense subspaces of X and E, respectively. Let

0 ^//Y ^ı ^//E1 π

//E ^// 0, and 0 ^//E ^ ^//E2 ̟

//X ^// 0

be the induced sequences, so that E1 = Z(Φ) and E2 = Z(Ψ) are the completions of Y ⊕^Φ E0 and E⊕^ΨX0, respectively, as explained in Section 2.3.

In this setting, we say that the concatenation Φ Ψ is trivial in Ext²(X, Y), and we write Φ Ψ∼0 for short, if the associated four-term sequence

0 ^//Y ^ı ^//E1 //

π◆◆◆◆''

◆◆ E2 ̟

// X ^//0

E ^

77♣

♣♣

♣

is trivial in Ext²(X, Y). We have the following operative version of Lemma 3.1. The proof is a simple adaptation that we leave to the patient reader.

Lemma 3.2. Let X, E, Y be quasi Banach spaces and Ψ :X₀ −→E andΦ :E₀ −→Y quasilinear maps such that Ψ[X₀]⊂ E₀, where X₀ and E₀ are dense subspaces of X and A, respectively. The following are equivalent:

(a) Φ Ψ∼0 in Ext²(X, Y).

(b) Φ has a quasilinear extension to E0⊕^ΨX0.

(c) Ψ has a quasilinear lifting to Y ⊕^ΨE0.

Our next task is to render these criteria manageable; with the same notation of the preceding Lemma, one has

Lemma 3.3. The following are equivalent:

(a) Φ Ψ∼0 in Ext²(X, Y).

(b) There is a homogeneous map H :X0 −→Y and a constant K such that kH(x+x^′)−H(x)−H(x^′)−Φ(Ψ(x+x^′)−Ψ(x)−Ψ(x^′))k ≤K kxk+kx^′k for every x, x^′ ∈X0.

Proof. (a) =⇒ (b) Let Γ : E0 ⊕^ΨX0 −→ Y be any quasilinear extension of Φ, so that Γ(e,0) = Φ(e) for every e∈E0. Define H(x) = Γ(Ψ(x), x) and check.

(b) =⇒ (a) Define Γ : E0⊕^ΨX0 −→ Y as Γ(e, x) = H(x) + Φ(e−Ψ(x)) and check that it is quasilinear. Obviously Γ(e,0) = Φ(e) for every e∈E0.

The function H will sometimes be called a witness that Ψ Φ∼0.

Locally convex twisted sums and K-spaces. The criteria provided by Lemma 3.1 and its relatives characterizes the triviality of spliced sequences in the category of quasi Banach spaces. The equivalence between (a) and (b) is true replacing quasi Banach spaces by Banach spaces and Ext²(X, Y) by Ext²_B(X, Y), and the same is true for quasi Banach modules. Parts (c) and (d) are more delicate since a quasilinear map acting between Banach spaces can well lead to a quasi Banach space which is not even isomorphic to a Banach space; see [25, 14]. One can isolate those quasilinear maps that, acting between Banach spaces, produce Banach spaces. This approach was pursued in [3]. Those

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subtleties are however unnecessary in our current circumstances. Let us explain why. A minimal extension of a quasi Banach space X is a short exact sequence of the form 0 −→ K −→ Z −→ X −→ 0. If all minimal extensions of X are trivial, that is, if Ext(X,K) = 0, then X is called a K-space. The spaces ℓp are K-spaces for all values of 0 < p≤ ∞ with the only exception of p= 1. Moreover, if X is a Banach K-space and one has an exact sequence 0 −→ Y −→ Z −→ X −→ 0 in which Y is a Banach space, then Z is isomorphic to a Banach space and so Ext(X, Y) = Ext_B(X, Y). The point of this discussion is to remark that when the spaces X and E are Banach K-spaces and Y is any Banach space, the space appearing in (b) is also necessarily a Banach space:

just look at the left descending sequence and then at the middle horizontal one. In this way, all the results of this Section remain true in the category of Banach spaces under the additional assumption that X and E are K-spaces.

4. The counterexample

Let X and Y be ℓ∞-modules and let A be any subset of ℓ∞. We say that a function f : X −→ Y commutes with A if f(ax) =af(x) for every x ∈ X and every a ∈ A. We are mostly interested in the following choices of A:

• The unitary group U ={u:N−→K such that |u|= 1}.

• The real unitary group UR={u:N−→R such that |u|= 1}.

Of course U = UR when the ground field is R. Note that if f : X −→ Y is a mapping acting between sequence spaces that commutes with U^R, then f preserves supports in the sense that supp(f(x)) ⊂ supp(x). Every centralizer defined on a sequence space is strongly equivalent to one that commutes withU.

Lemma 4.1. Let X and Z be quasi Banach sequence spaces andY a Banach sequence space. If Ψ : X⁰ −→ Z and Φ : Z⁰ −→ Y are centralizers which commute with UR and Φ Ψ∼0in Ext²(X, Y)then one can choose a witness function H that commutes withUR. Proof. If H : X⁰ −→ Y is as in Lemma 3.3(b), then for every unitary u, and in particular foru∈UR one has

ku⁻¹ H(u(x+x^′))−H(ux)−H(ux^′)

−Φ(Ψ(x+x^′)−Ψ(x)−Ψ(x^′))k ≤K kxk+kx^′k . If we identify the groupURwith the “Cantor group”{1,−1}^N, it is clear that the product topology of the later corresponds to the relative weak* topology ofURwhenℓ∞is regarded as the dual of ℓ1. In particular UR is a compact group for the weak* topology. Let du denote the Haar measure on UR. Observe that for finitely supported x, the “orbit”

URx= {ux: u∈ UR} is finite and so, the mapping u∈ UR 7−→u⁻¹H(ux)∈ Y is weak*

to norm continuous. Define a new mapping He :X⁰ −→Y by the Bochner integral H(x) =e

Z

U_R

u⁻¹H(ux)du

and check. Note thatH(x) agrees with the average ofe u⁻¹H(ux) over those real unitaries

usuch that u= 1 off supp(x).

We now specialize to X =Z =Y =ℓ2 and make a careful using of the symmetries of the Kalton-Peck centralizer Ω.

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Corollary 4.2. Let Φ : ℓ⁰₂ −→ ℓ₂ be centralizer commuting with UR. If Φ Ω ∼ 0, then there exist a homogeneous, support preserving H :ℓ⁰₂ −→ ℓ2 and a constant K such that, wheneverx, y ∈ℓ⁰₂ are disjoint and of equal norms,

kH(x+y)−H(x)−H(y)−Φ(x+y)k2 ≤Kkxk2. Proof. Ifx, y are disjointly supported and of equal norm in ℓ2, then

Ω(x+y)−Ω(x)−Ω(y) = log 2

2 (x+y).

The result follows using the homogeneity of Φ andH.

In order to exploit the estimate provided by the preceding result we need to select pairs of disjoint sums of vectors from the unit basis of ℓ⁰₂ in a judicious way. This is achieved through certain partitions, defined below. The idea is quite simple:

• Fix k∈N and start with the set 2^k (that is, a set of cardinality 2^k).

• Split it into two halves with the same number of elements. Then split the resulting sets into two halves and continue until reaching the singletons.

• Don’t forget to keep track of the whole process labelling all the sets.

We can formalize this procedure by using a dyadic tree of finite height. Let T_k be the dyadic tree of heightk whose elements are words of length at most k written with 0s and 1s, including the “empty word”, which has length 0. Given α = α1α2· · ·αn in T_k, with n < k, we putα0 =α1α2· · ·αn0 and α1 =α1α2· · ·αn1.

Definition4.3.An adequate partition of 2^kis a set-valued functionI :T_k−→P(2^k) such that:

• I(∅) = 2^k, I(α) is nonempty for every α∈T_k.

• If α has length less than k, then I(α) is the disjoint union of I(α0) and I(α1).

The setsI(α) forαof fixed length 0≤n ≤kform a partition on 2^kinto 2ⁿmany subsets of cardinality 2^k−n. It takes only a moment’s reflection to realize that an adequate partition is essentially a linear order on 2^k. Indeed for every adequate partition there exist a unique order4 such thatx4y whenever x∈I(α0) and y∈I(α1). This correspondence will be used later.

Each adequate partition on 2^k gives rise to a family of vectors ofℓ2(2^k), parametrized by α∈T_k just taking

x_α= 1_I(α) = X

i∈I(α)

e_i.

Lemma 4.4. Let Φ : ℓ⁰₂ −→ ℓ2 be centralizer commuting with UR. If Φ Ω ∼ 0 then there is a constant K = K(Φ) such that if S is a subset of N with |S| = 2^k and I, J : T_k −→P(S) are adequate partitions, then

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X

α∈T_k

Φ(xα)−Φ(yα)

2 ≤2Kk√ 2^k−1,

wherex_α andy_α are the vectors associated to I andJ, respectively, that is,x_α =P

i∈I(α)e_i and yα =P

i∈J(α)ei for α∈T_k.

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Proof. Let H and K be as in Corollary 4.2. Since x₀ and x₁ have the same norm and x0+x1 =x∅ = 1S we have

kH(x∅)−H(x₀)−H(x₁)−Φ(x∅)k2 ≤K√ 2^k−1. The same estimate holds using y0 and y1. Since y∅ =x∅ we get (11) kH(x0) +H(x1)−H(y0)−H(y1)k² ≤2K√

2^k−1. We claim that, for every 0< j≤k,

(12)

X

|α|=j

H(xα)−H(yα)

+ X

|α|<j

Φ(xα)−Φ(yα)

2 ≤2Kj√ 2^k−1,

where|α|denotes the length of α. Indeed, when j = 1 this reduces to (11). Now assume that (12) holds for j =i−1 and let us check it for j =i.

By the induction hypothesis this comes from

X

|α|=i

(Hxα−Hyα)− X

|γ|=i−1

(Hxγ−Hyγ)− X

|δ|<i−1

(Φxδ−Φyδ) + X

|α|<j

(Φxβ −Φyβ)

2

≤ X

|γ|=i−1

H(xγ0) +H(xγ1)−H(xγ0+xγ1) + Φ(xγ0+xγ1)

2

+ X

|γ|=i−1

H(yγ0) +H(yγ1)−H(yγ0+yγ1) + Φ(yγ0+yγ1)

2

which using that H and Φ preserve supports and that

kH(zα0) +H(zα1)−H(zα0+zα1) + Φ(zα0+zα1)k ≤K√ 2^k−j, forz =x, y, gives an upper estimate of

2√

2^j−1K√

2^k−j = 2K√ 2^k−1

and proves the claim. When j =k then allxα and yα are single vectors of the basis, the first summand in (12) is null, and

X

α∈T_k

Φ(xα)−Φ(yα)

2 = X

|α|<k

Φ(xα)−Φ(yα)

2 ≤2Kk√ 2^k−1.

Let us say that centralizer Φ :X −→Y acting between symmetric sequence spaces is symmetric if, for every permutation σ :N −→N, one has Φ(x◦σ) = (Φ(x))◦σ. This is not the traditional definition, but every symmetric centralizer in the traditional sense is strongly equivalent to one of this form.

Note that if Φ is a symmetric centralizer, then the left-hand side of (10) is 0; indeed in that case there there are scalars λj for 0 ≤ j ≤ k such that Φ(xα) = λ|α|xα and Φ(yα) = λ|α|yα for all α ∈ T_k. So, in order to obtain our counter-example we need a highly nonsymmetric centralizer.

Let us consider the following finite-dimensional versions of the centralizer Ω defined ine Section 2.5. Fix m, n∈N and partition the productm×n into m subsets of cardinality n as follows

Ai ={(i, j) : 1≤j ≤n} (1≤i≤m).

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Then defineΩem,n :ℓ2(m×n)−→ℓ2(m×n) by Ωe_m,n(x) = X

1≤i≤m

x_ilog kxk kxik,

where xi = x1Ai. By Lemma 2.3 we have C(Ωem,n) ≤ C(Ω) for all m, n. Note that if x has exactlyq nonzero chunks and they all have the same norm, then

Ωe_m,n(x) = log(q)x.

Fix now some k and identify 2^k with the product 2^r×2^s with k =r+s. Let 4 be the associated lexicographic order, i.e., for (a, b),(c, d)∈2^r×2^s,

(a, b)4(c, d)⇐⇒a < c or (a =c and b≤d).

Let I : T_k −→ P(2^k) be the adequate partition associated to 4. We also consider the lexicographic order “symmetric” to 4, namely

(a, b)4^′ (c, d)⇐⇒b < d or (b =d and a≤c).

LetJ be the partition induced by 4^′. Now we follow the notations introduced just after Definition 4.3, in particular

xα = X

i∈I(α)

ei, yα = X

j∈J(α)

ej (α∈T_k).

Lemma 4.5. Let k =r+s given, I and J as before, with associated vectors (xα) and (yα), respectively. Let Ωe be the centralizer Ωe2^r,2^s :ℓ2(2^r×2^s) −→ℓ2(2^r ×2^s). Then, for every α∈T_k one has

|α| ≥r =⇒ Ω(xe _α) = 0,

|α|< r =⇒ Ω(xe α) = (r− |α|) log(2)xα,

|α| ≥s =⇒ Ω(ye α) = (k− |α|) log(2)yα,

|α|< s =⇒ Ω(ye _α) =rlog(2)y_α. In particular

X

α∈T_k

Ω(xe α)−Ω(ye α)

2 = log(2)rs√ 2^k.

Proof. We suggest the reader to take a pencil and scribble some trees and rectangles.

It is clear from (5) that ifxhas a single nonzero chunk, thenΩ(x) = 0, so certainlye Ω(xe α) = 0 if|α| ≥k−s=r. On the other hand, if|α|< r, thenx_α is the sum of 2^r−|α|many chunks of equal norm and so Ω(xe α) = (r− |α|) log(2)xα. Regarding yα, for |α| ≥k−r =s, it is a sum of unit vectors in 2^k−|α| disjoint Ai’s, so Ω(ye α) = (k− |α|) log(2)yα and if |α|< s it is the sum of equal norm vectors in all the 2^r sets Ai’s, so Ω(ye α) = rlog(2)yα.

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Finally we compute P

α∈T_k Ω(xe α)−Ω(ye α)

log 2 = X

|α|<r

Ω(xe _α)− X

|α|≥s

Ω(ye _α)− X

|α|<s

Ω(ye _α)

= X

|α|<r

(r− |α|)xα− X

|α|≥s

(k− |α|)yα− X

|α|<s

ryα, which, taking into account that for each 0≤n ≤k

X

|α|=n

xα = X

|α|=n

yα=x∅ =y∅ = X

i∈2^r×2^s

ei, is equal to

−r(s−1)−X

n≤r

n+X

n<r

n

x∅ =−rsx∅

and its norm is rs√

2^k.

Theorem 4.6. Let(Ai)be a partition of N. Let Ω :ℓ⁰₂ −→ℓ2 be the Kalton-Peck map and letΩe be the centralizer associated to (Ai), as in Section 2.5. If for everyn there exist nsets in(Ai)whose cardinality is at leastn, thenΩ Ωe ≁0. In particular, Ext²(ℓ2, ℓ2)6= 0. Proof. For each k we can find a subspace generated by 4^k vectors of the basis and where Ω identifies with the centralizere Ωe₂^k_,2^k as described in Lemma 4.5 for r = s = k.

Then the two previous Lemma together with the symmetry of Kalton-Peck map yield that if Ω,e Ω∼0 and K is the associated constant then one gets the contradiction

log(2)k²2^k = X

α∈T_k

Ω(xe α)−Ω(ye α)

2 ≤2√

2Kk2^k.

Since the extension induced by Ω Ω lives ine Band is not trivial in Q one gets

Corollary 4.7. Ext²_B(ℓ2, ℓ2)6= 0.

5. Miscellaneous applications

The remainder of the paper is devoted to presenting a number of applications of the main result. These range from classical operator theory to Banach modules, including some issues about the difference between Ext² and Ext²_B.

5.1. Spinning around Hilbert space. Hilbert space is a central object of Banach space theory in many respects. The ensuing application exploits this fact in a rather direct way. A Banach spaceX containsℓⁿ₂ uniformly complemented if there is a constant C such that, for everyn ∈Nthere are operatorsI :ℓⁿ₂ −→X and P :X −→ℓⁿ₂ such that P I is the identity on ℓⁿ₂ and kIkkPk ≤C. This property is “self-dual” (X has it exactly when X^∗ does) and weaker than B-convexity (= nontrivial type p >1).

Corollary 5.1. If X and Y are Banach spaces containing ℓⁿ₂ uniformly comple- mented, then Ext²_B(X, Y)6= 0.

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Proof. We write the proof assuming X and Y separable. The general case does not present any additional difficulty. Given a (separable) Banach spaceX we fix an isometric quotient map Q1 : ℓ1 −→ X and we set K¹(X) = kerQ1. Then we fix a quotient Q2 : ℓ1 −→ K¹(X) and set K²(X) = kerQ2. Splicing through K¹(X) we obtains the exact sequence

0 ^// K²(X) ^//ℓ1 //

Q◗◗₂◗◗◗◗((

◗ ℓ1

Q1

//X ^//0

K¹(X)

66

♠♠

♠

that allows us to “reduce” the length of extensions. Indeed, for all Banach spacesY, one has Ext²_B(X, Y) = Ext_B(K¹(X), Y). This follows from the fact that ℓ1 is projective in the category of Banach spaces. If we specialize to the case where X =H is a separable Hilbert space, we get Ext²_B(H, Y) = Ext_B(K¹(H), Y) for all Y. Now, by the main result Ext²_B(H, H) = Ext_B(K¹(H), H) is nonzero and a uniform boundedness argument (cf. [4, Theorem 2]) yields Ext_B(K¹(H), Y)6= 0. We have thus arrived at:

Claim. IfY contains ℓⁿ₂ uniformly complemented, then Ext²_B(H, Y) 6= 0. In particular, Ext²_B(H, X^∗)6= 0.

In view of the duality formula Ext²_B(E, F^∗) = Ext²_B(F, E^∗), valid for all Banach spaces E and F, one also has Ext²_B(X, H^∗) 6= 0. But since H^∗ is isometric to H this means that ExtB(K¹(X), H) 6= 0 and applying again [4, Theorem 2] we obtain Ext²_B(X, Y) =

ExtB(K¹(X), Y)6= 0.

5.2. Impossibility of extending certain operators. Kalton shows in [18] that the Kalton-Peck spaces Zp(ϕ) for 1 < p < ∞ have the remarkable property that every C(K)-operator defined on any subspace ofZp(ϕ) extends to an operatorZp(ϕ)−→C(K).

Curiously, one has:

Corollary 5.2. Let X be a separable Banach space containing ℓⁿ₂ uniformly com- plemented. There exists an embedding u : X −→ C[0,1] and an operator v : ℓ2 −→

C[0,1]/u[X] that cannot be extended to Z₂.

Proof. We will make the proof forX =ℓ2 and leave the reader to derive from that and the Claim above the general case. Let u : ℓ2 −→ ℓ∞ be an isomorphic embedding.

Since ℓ∞ is an injective Banach space one has a (pull-back) commutative diagram 0 ^//ℓ2 u

//ℓ∞ //ℓ∞/u[ℓ2] ^//0

0 ^//ℓ2 // Ze2 //

V

OO

ℓ2 v

OO //0

where V is an extension of u and v is the factorization of V through the quotient map.

The (self-adjoint, if K = C) subalgebra of ℓ∞ generated by V[Ze2] is a Banach space